If $f(0)=0 ,f(1)=1 ,f'(0)=f'(1)=0$, then $|f''(x)|>4$

Question

Let $ f $ be a twice differentiable function with $ f (0) = 0, f (1) = 1$ and $ f '(0) = f' (1) = 0 $, then $ 4 \leq | f'' (x ) | $, for some $ x \in [0,1] $.

I tried using the mean value theorem for derivatives with

$ \dfrac{f '(1)-f' (0)}{f (1)-f (0) } = f '(c) = 0$

so there is a value at which $ f ''(c) = 0 $, but that did not help me.

Any ideas? Any help is appreciated.

Also, kEoz, feel free to upvote answers to your questions. – Matthew Conroy Dec 29 '13 at 21:39 — Matthew Conroy, Dec 29 '13 at 21:39
sorry i dont know that y read about this – kEoz Dec 29 '13 at 21:55 — kEoz, Dec 29 '13 at 21:55

score 5 · Answer 1 · answered Dec 29 '13 at 21:27

Let $g(x)=f'(x)$. Then we know $\int_0^1 g(x)\,dx=1$ and $g(0)=g(1)=0$.

If $|g'(x)|<4$ everywhere, then due to the MVT $g(x)$ must be smaller than $4x$ and also smaller than $4(1-x)$. So $g$ is entirely within the triangle with corners $(0,0)$, $(\frac 12,2)$, $(1,0)$. Can its integral then be large enough?

If $f(0)=0 ,f(1)=1 ,f'(0)=f'(1)=0$, then $|f''(x)|>4$

1 Answers1